Pascal's Theorem

We are given five points P, P', Q, R, and
S, and can show that the conic lying on
these five points was given by the locus of blue points.

Now let us define N as the intersection of x and
z. We see in the diagram that N is on the conic, and
can verify that our construction would send PN to P'N.
We can state this as a theorem:

If PR.QN, RP'.NS, and P'Q.SP are collinear, then
N lies on the conic determined by PP'QRS.

Rather than saying that N lies on the conic determined by
PP'QRS, we could simply say that NPP'QRS lie on a conic. It
will also simplify things to speak about the hexagon PRP'QNS;
then the points lie on a conic if and only if the hexagon is
inscribed in that conic. Making these modifications and some
changes of labelling, we have the theorem:

If opposite sides of a hexagon (ABCDEF) intersect in three points
(AB.DE, BC.EF, CD.FA) which are collinear, then the hexagon may
be inscribed in a conic.

This is known as the converse of Pascal's theorem.

So Pascal's theorem says:

If a hexagon (ABCDEF) is inscribed in a conic, then opposite
sides intersect in three points (AB.DE, BC.EF, CD.FA) which are
collinear.

Proof: Define L=BC.EF, M=CD.FA, N=AB.DE; we want
to show L, M, N collinear.
To do this, we also define J=AB.CD, K=FA.BC. Then

( A, N; J, B) =
(DA,DE;DC,DB) =
(FA,FE;FC,FB) =
( K, L; C, B).

Thus the projectivity sending A to K, J to
C, and B to B must also send N to
L. Since that projectivity is the perspectivity defined
by M, M sends N to L; in other words,
L, M, and N are collinear as desired. We may say
that they lie on the Pascal line of the hexagon.

Note that Pascal's theorem is true regardless of where the
points lie on the conic. The diagram above shows a very
non-convex hexagon, but since projective geometry does not
deal with convexity, a convex hexagon would do just as well.
However, a convex hexagon might have a Pascal line too far off
the diagram to be seen.